In the xy-plane, the graph of a cubic polynomial function p with real coefficients has exa… — SAT

In the $xy$ -plane, the graph of a cubic polynomial function $p$ with real coefficients has exactly two $x$ -intercepts, at $(1, 0)$ and $(4, 0)$ . If the graph of $p$ passes through the points $(0, -8)$ and $(2, 2)$ , what is the value of $p(6)$ ?

Answer: 10

Answer

A cubic polynomial with real coefficients and exactly two

x

-intercepts at

(1, 0)

and

(4, 0)

must have one root of multiplicity 1 and one root of multiplicity 2. This yields two possible forms:

p(x) = a(x - 1)^2(x - 4)

p(x) = a(x - 1)(x - 4)^2

. Substituting the

y

-intercept

(0, -8)

into the first form gives

a = 2

, but the resulting polynomial

p(x) = 2(x - 1)^2(x - 4)

does not pass through

(2, 2)

since

p(2) = -4

. Substituting

(0, -8)

into the second form gives

a = \frac{1}{2}

, and the resulting polynomial

p(x) = \frac{1}{2}(x - 1)(x - 4)^2

correctly passes through

(2, 2)

since

p(2) = 2

. Finally, evaluating this function at

x = 6

yields

p(6) = \frac{1}{2}(6 - 1)(6 - 4)^2 = 10

Step-by-Step Solution

Determine the possible forms of the cubic polynomial based on its

x

-intercepts.

Two possible cases: Case 1:

p(x) = a(x - 1)^2(x - 4)

or Case 2:

p(x) = a(x - 1)(x - 4)^2

, where

a

is a non-zero constant.

Since the polynomial has degree 3 and exactly two distinct

x

-intercepts, one intercept must be a root of multiplicity 2 (tangent to the

x

-axis) and the other must be a root of multiplicity 1 (crossing the

x

-axis).

Solve for the constant

a

in Case 1 using the

y

-intercept

(0, -8)

a = 2

, yielding the candidate function

p(x) = 2(x - 1)^2(x - 4)

Substituting

x = 0

and

p(0) = -8

into the equation for Case 1 allows us to solve for

a

Test the point

(2, 2)

in the Case 1 candidate function.

p(2) = -4 \neq 2

, which means Case 1 is incorrect.

The correct function must satisfy all given points, including

(2, 2)

Solve for the constant

a

in Case 2 using the

y

-intercept

(0, -8)

a = \frac{1}{2}

, yielding the candidate function

p(x) = \frac{1}{2}(x - 1)(x - 4)^2

Substituting

x = 0

and

p(0) = -8

into the equation for Case 2 allows us to solve for

a

Test the point

(2, 2)

in the Case 2 candidate function.

p(2) = 2

, which matches the given point.

Confirming that the Case 2 function is consistent with the point

(2, 2)

Evaluate the correct function at

x = 6

p(6) = 10

Substituting

x = 6

into the verified function

p(x) = \frac{1}{2}(x - 1)(x - 4)^2

to find the final answer.

Key Concept

Identifying the algebraic form of a polynomial from its

x

-intercepts and multiplicities, and determining unknown coefficients using coordinate points.

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